1已知a1 =2 点an+1=2a+3.求数列an的通项公式 2已知数列an中a1=1.an+1=an+2n,秋通项

1．已知a1=2 点an+1=2an+3.求数列an的通项公式,
2．已知数列an中a1=1.an+1=an+2n,求通项公式,
解1．由an+1=2an+3得an+1+3=2an+6=2(an+3).设bn=an+3,则有
bn+1=2bn,故{ bn }是公比为q=2的等比数列,b1=a1+3=2+3=5,由等比数列通项公式得bn=b1q^(n-1)= 5×2^(n-1),故
an=bn-3=5×2^(n-1)-3
2.由an+1=an+2n得an+1-an=2n,于是有
a2-a1=2
a3-a2=4
a4-a3=6
….
an-an-1=2(n-1)
上面所有式相加得
an- a1=2(1+2+…+(n-1))=2×n(n-1)/2=n(n-1)
an = a1+n(n-1)=n(n-1)+1